At what angle should a body be projected with a velocity 24 ms⁻¹ just to pass over the obstacle 14 m high at a distance of 24 m. [Take g = 10 ms⁻²]
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Question

At what angle should a body be projected with a velocity 24 ms⁻¹ just to pass over the obstacle 14 m high at a distance of 24 m. [Take g = 10 ms⁻²]

G1 Physics · Accepted Answer

For a projectile to pass through (x, y) = (24 m, 14 m) with u = 24 m/s and g = 10 m/s², use y = x tanθ - gx²[1 + tan²θ]/(2u²). Let p = tanθ. Then 14 = 24p - 5(1 + p²), so 5p² - 24p + 19 = 0. This gives p = 1 and p = 19/5. Thus both tanθ = 1 and tanθ = 19/5 make the projectile just pass the obstacle, corresponding to options A and B. The printed source key lists only A.